$\overline{AB}$ = $4\sqrt{5}$ $\overline{BC} = {?}$ $A$ $C$ $B$ $4\sqrt{5}$ $?$ $ \sin( \angle BAC ) = \frac{2\sqrt{5} }{5}, \cos( \angle BAC ) = \frac{ \sqrt{5}}{5}, \tan( \angle BAC ) = 2$
Solution: $\overline{AB}$ is the hypotenuse $\overline{BC}$ is opposite to $\angle BAC$ SOH CAH TOA We know the hypotenuse and need to solve for the opposite side so we can use the sine function (SOH) $ \sin( \angle BAC ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{\overline{BC}}{\overline{AB}}= \frac{\overline{BC}}{4\sqrt{5}} $ $ \overline{BC}=4\sqrt{5} \cdot \sin( \angle BAC ) = 4\sqrt{5} \cdot \frac{2\sqrt{5} }{5} = 8$